The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $22.6$ years; the standard deviation is $4.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $27.4$ years.
Solution: $22.6$ $17.8$ $27.4$ $13$ $32.2$ $8.2$ $37$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $22.6$ years. We know the standard deviation is $4.8$ years, so one standard deviation below the mean is $17.8$ years and one standard deviation above the mean is $27.4$ years. Two standard deviations below the mean is $13$ years and two standard deviations above the mean is $32.2$ years. Three standard deviations below the mean is $8.2$ years and three standard deviations above the mean is $37$ years. We are interested in the probability of a gorilla living less than $27.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the gorillas will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $17.8$ years and the other half $({16\%})$ will live longer than $27.4$ years. The probability of a particular gorilla living less than $27.4$ years is ${68\%} + {16\%}$, or $84\%$.